 We had started discussing the concept of four vectors and in our last lecture, we started talking about the velocity four vector. We realized that if you take the first three components of standard velocity vector which we call as ux, ui and uz, these do not form the first three components of velocity four vector. The reason for that is that in order to evaluate velocity, we have to divide the displacement by time or other time interval. Now, though the displacement does form components of the four vector, the time does not because delta t is a frame dependent quantity. So, we agreed that in order that we want to construct a velocity four vector, we should not divide by delta t, but should divide by delta tau the proper time interval between the displacement between the two events of the displacement of the particle. Once we do that, then the standard values of velocity components ux, ui, un and uz get multiplied by gamma u. They form the first three components of velocity four vector. And then we also took one example and showed that how to construct a velocity four vector and also showed that once the same particle is looked by another observer in a different frame, then how this velocity four vector will transform using the standard transformation equation. So, this is the recapitulation of what we had done. We described velocity four vector, then eventually we also derived velocity transformation using velocity four vector. And as I have mentioned, then we gave an example. Let us recall that our idea, the way we started looking about the four vectors was eventually to define a momentum four vector. So, now, let us start looking at a new definition of momentum. We have talked about displacement, then we talked about the velocity, then let us talk about the momentum. So, we now start talking about the momentum four vector and try to see the way we have to look for a new definition of momentum. Now, in order that we look into a new definition of momentum, I have to multiply this particular thing by something which has a dimension of mass or we have to multiply a velocity four vector with something which has a dimension of mass. And this particular mass should be frame independent. We do not normally bother about in the classical mechanics about mass which we always take this to be frame independent. But let us be specific and in order to convert from velocity four vector to momentum four vector, let us define another quantity which we call as a rest mass of the particle. So, in view of the fact that there is some universality associated with the concept of four vectors, the idea is that if we define, we define this particular four vector from the momentum point of view, then it would be easier to see that its conservation is obeyed universally, it means in all the rational frame of reference. So, as I said, let us define a proper or what we call as a rest mass which we call as M naught. And to be specific, let us say that this mass is the mass of the particle which is determined in a frame of reference in which this particle is at rest. So, that is the reason popularly this is always called rest mass of the particle. And because this is something which is always measured in a frame in which the particle is at rest, this is a four scalar. So, even if I change, go to different frame of reference, M naught, the rest mass of the particle should not change or does not change. So, we define a four scalar which is a frame independent quantity called rest mass of the particle calling this as M naught. Once I multiply this M naught by the velocity four vector, I am expected to get something which has dimension of momentum and let us start calling that as momentum four vector. We will eventually see how to physically interpret and how to see that conservation of momentum becomes a universal principle. So, what I do? We had already taken the velocity four vector u and the components of that particular four vector was gamma u multiplied by u x, u y, u z, i c. It means the first component is of velocity four vector was gamma u u x, second component was gamma u u y, the third component was gamma u u z, the fourth component was gamma u i c. This entire thing I have multiplied by M naught and called this as a momentum four vector. Again I repeat this M naught is supposed to be frame independent quantity called the rest mass or the proper mass of the particle. So, let us assume at the moment that the first three components of this particular momentum four vector the way we have defined are the new definitions of momentum. If we assume that let us see whether the conservation law really becomes universal because that is what is our idea. That is what we are aiming at that the conservation of momentum principle becomes a universal principle. It means if it is obeyed in one frame of reference it must be obeyed in all other frames of references. So, let us try to look whether this definition can yield me that particular universality. Before we do that let us write these components. If you remember we had written the velocity four vector as gamma u u x that was the first component, the second component was gamma u u y, the third component was gamma u u z and the fourth component was gamma u i c. I have multiplied this by M naught. So, my first component of momentum four vector becomes M naught gamma u u x, M naught gamma u u y, M naught gamma u u z and the fourth component is gamma u M naught gamma u i c. As we had agreed that the first three components of this are the first three components of momentum, the new definition of momentum as per special theory of relativity. It would mean that I would call this quantity here S p x which is the x component of the momentum. This quantity I will call S p y, this quantity I will call S p z and this fourth quantity at the moment we have to find out what it is. Let us suppose this fourth component I am just writing this as A 4. We will look at this particular A 4 little more clearly in the later part of this particular lecture. So, this is what has been written in this particular transparency that p I am writing in the form of p x, p y, p z and A 4 while p x is equal to M naught gamma u u x, p y is equal to M naught gamma u u y, p z is equal to M naught gamma u u z and A 4 something which we do not know is M u gamma u i c. These three I will call as the x, y and z components of the momentum of the particle. So, if we assume this particular thing let us see whether conservation of momentum would be universally accepted phenomenon. It means all the observers in all different frames will find the conservation of momentum law to be holding with these new definitions of momentum. What is conservation of momentum? The conservation of momentum tells that if you have a system of particles and if they interact and if there is no external force on the system of the particle. In fact, we differentiate between internal forces and the external forces. So, if we have a system of n particles and each are interactive with each other and therefore, each is influencing applying a force on other particle. So, long the force is being applied on a particle by another particle of the system that particular force I will call as an internal force. But any force which is being applied on any of the particle by something which is not a part of the system we call that as an external force. So, if you have a large number of particles interacting something like gas molecule you have one particle here you have one particle here one particle here one particle here they keep on coming somewhat collides here something collides here this could be for example, that matter even charged particles. So, there could be electrostatic force between these two electrostatic force between these two between these these. So, they are all interacting, but if I call this as my system any force which is caused by particle within the system that is what we will call as internal force. But assume that this whole thing is kept under gravitational field and there is an earth here which is attracting all these particles and this earth is not a part of the system then that particular force will be external force to the system. So, conservation law conservation of momentum law says that if external force is zero then the net vector momentum of all the particles must always remain constant that is what is called conservation momentum momentum will be conserved. So, if no external force exist on the system then the total momentum the vector sum of all the moment momentum of all the particles must remain identical it should not change. So, this is what I have shown in this particular picture in which I have shown only two particles. So, let us suppose this is one particle there is another particle these two particles interact and in this particular case I have assumed that they just collide collision is actually one of the form of interactions. So, once they collide this particular particles velocity may change which I have not shown it is not important for our argument here. Similarly, the velocity of this particular particle is also likely to change after the collision we have normally seen the collisions of particles collisions of billiard balls. So, many collisions these are very natural things collision is a very standard classical mechanic phenomenon. Now, what conservation of momentum would mean that if I take the initial momentum of the first particle which I am writing as p 1 i and the initial momentum of the second particle which I am writing as p 2 i even though their momentum would change after collision I must have p 1 f if p 1 f is the final momentum of the first particle plus p 2 f where p 2 f is the final momentum of the second particle because there is no external force to the system therefore momentum must be conserved therefore according to conservation of momentum law this initial momentum must be equal to the final momentum. Now, what is our requirement? Our requirement is that if the same collision is being seen by some other observer in some different frame he may find different value of initial momentum of the two particles but whatever he finds he or she finds immediately after the collision the final momentum in his frame or her frame should also be the same. So, though they may disagree on the value of the momentum like they disagree on the value of the velocity but they should not disagree on conservation of momentum initial momentum must be final momentum whatever is that value of initial momentum may be that is what is conservation of momentum and that is what I wanted. To happen with this new definition of momentum we have realized earlier that if we just use Lorentz transformation and do not change any other definition of momentum then this will not be the case we had given an example earlier. So, this is what I have written here. Universality of conservation of momentum implies that in any other frame even though individual momentum may be different but still we must have p 1 i in fact I have put a prime here to show that this is a different frame plus p 2 i prime this could be different from p 1 i could be different from p 2 i because they are primed. So, the momentum individual momentum value may be different in different frame in fact even this sum may be different in different frame but whatever might be this value immediately after the collision that particular observer sitting in this particular frame of reference must also conclude that the final momentum of the first particle plus final momentum of the second particle must be equal to the initial momentum of the first particle plus initial momentum of the second particle. This is what he should also conclude he or she should also conclude that conservation of momentum is valid in his or her frame also. Now as per new definition of momentum in fact we have given a definition of new definition of momentum and we want to see whether this obeys this universality condition. Then as per the new definition of the momentum the first three components of the vector four vector r x y and z component of momentum vector that is what we have discussed. Hence the sum of the three components of individual momentum would transform as follows and now I make it general instead of talking two particles I talk of n particles. And remember when I am talking of conservation of momentum as a vector quantity it means all the individual x y and z components must also be conserved. A vector does not necessarily mean that its only magnitude should be same it also means its direction should be same. It means its x y and z components individually must be conserved. So even the summation of x value of momentum summation of y component of momentum summation of z component of momentum should also conserved. Now let us look at the transformation equation because we have started with the four vector momentum concept. Let us see whether this universality of conservation law will hold good with this new definition of momentum or not. So this looks a somewhat a complicated equation on this transparency but let us try to spend a little bit of time to understand it. So I have assumed that my first three components of the momentum are p x p y and p z component utilizing the new definition of momentum. Now I assume that they are n particles and initially at a time when I start observing I find that the x component of the first particle is p x 1 of the second particle is p x 2 for the third particle is p x 3 and this i signifies that these are the initial values at the time when you started looking at the observation. So if I take for all the n particles I have to sum at all those values so I will get summation of k over 1 to n p x k i where k would vary from 1 to n because they are n particles. So this will give me the sum of the x component of momentum for all the particles. Similarly this will give me the sum of the y component of the momentum of all the particle. This will give me the sum of the z component of all the particles, z component of momentum of all the particles. A4 as of now I do not know what it is. So I will just say A4 k i of course I know the value of A4 in terms of gamma u I am not I c that is a different question but the physical as of now we do not know what does physically it represent and what is its role as far as conservation of momentum is concerned. Now we had earlier agreed that if p x, p y, p z and A4 are the components of a four vector if I take two four vector and add them they will also be the components of the four vector. It means these summations must also transform by the same transformation equation that we have written earlier. It means if I go to a different frame of reference which I call as s frame of reference look again at all those particles and find that in s frame of reference the summation of x component of momentum is given by this y by this z by this and A4 by this then if I open up this equation that should give me this value sum of the x component of the momentum in s frame of reference in terms of this particular initial values of some momentum in s frame of reference. This is the transformation equation which is has to be used to transform the initial value of momentum to the initial value of momentum in s frame of reference using the transformation equation. Let me just expand this matrix it is rather easy to expand we have done some examples earlier so let us not spend too much of time here what I will do here I will just expand this particular thing here this should be equal to gamma times this particular expression plus 0 times this plus 0 times this plus i beta gamma times this this is equal to this as we have seen it earlier number of times this is equal to this again as we have seen number of times now as far as the fourth component is concerned this will be minus i beta gamma times this plus 0 times this plus 0 times this plus gamma times this. So, I am just expanding it in the next transparency to write these things into four different equations this is what it is. So, some of the x component of the momentum of all the particles in s frame of reference would be given by this particular expression which is here summation of k from 1 to n p prime because this s frame of reference x k i k varies from 1 to n should be equal to gamma times whatever was the value of this corresponding quantity in s frame of reference plus i beta gamma times summation of k is equal to 1 to n a 4 k i whatever this a 4 b these were state forward equation now we also see the transformation of a 4 prime this is equal to minus i beta gamma times the first component plus gamma times the fourth component. So, these are the equations that we get after expanding that particular matrix or expanding that particular matrix equation. Now, as far as these two things are concerned these are straightforward let us look little more closely at this what is conservation momentum imply that in s frame of let us assume that in s frame the momentum is conserved. If s frame the momentum is conserved it means after the interaction you know they are all colliding with each other hitting each other and after some time you will find that the individual value of momentum of all of them have changed, but because there is no external force to the system which we are assuming then the sum of the final value of the momentum the x component must be same as the initial value similarly for y similarly for z. Now, I want this conservation I assume that this conservation law is holding good in s frame now I want that when I transform these things to s frame of reference there also the conservation of momentum must be holding good. So, this is given to me that this is equal to this the initial value is equal to final value which will be the case if there is no external force to the system similarly this is equal to this this is equal to this and I want to have a case where in s frame of reference also equivalent equations hold good with p b replaced by p prime. Now, out of this as I have said x equation is much more this is what we have written as we want that this also to be true in a different frame we must have p prime x i should be equal to p prime should be p prime should be equal to p prime y component p prime z component should be equal to p prime z component this is what I have been saying. Let us look at the first equation which is much more critical because for the y components they are equal anyway. So, this is the first equation which I have picked up from the transformation equation which tells me what is the sum of the x component of the momentum in s frame frame of reference where I transform this when I know these values how they will transform to s frame frame of reference and as we can see very easily that this summation would be equal to gamma times this summation plus i beta gamma times summation of this a 4 whatever this a 4 b. Now, we have said that after collision or after interaction the particles have all collided with each other and after some time we are observing then conservation of momentum is obeyed good in s frame it means the final value must be the same as this particular value. So, once interaction has taken place I have been given that this remains unchanged after collision this does not change after multiple collision there could be multiple collisions with individual particles but this does not change. Now, using this particular value I have obtained that this will be the value of the sum of the momentum in s frame of reference and I want that after the interaction because this has not changed this should also not be changed but I realize that this quantity does not solely depend on this but it also depends on a 4. If it so happens that after interaction this does not change this summation does not change but this summation changes then I will not be assured of conservation of x component because p prime would then change summation of p prime would change it means after the interaction the value of p prime x summation of x component of p prime would be different from the initial value and that is what we do not want it means if I want that this particular conservation should be holding on in s frame of reference I must look much more carefully at this particular value here. If this value also does not change after the interaction then only it is possible that this particular value will also not change after the interaction hence it is extremely critical to know what is this particular a 4 component and only if this particular a 4 component is conserved then only I am guaranteed that the momentum will be conserved in all other frames. So, we thus realize that the universality of conservation of momentum critically depends on the conservation of the fourth component of 4 vector. Now, if we go back to our thinking of our classical mechanics we have two specific conservation laws one is what is called conservation of momentum another is conservation of energy. However, in classical mechanics we have come across situations where what we call as mechanical energy need not be conserved. We have given an earlier example of completely inelastic collision where two particles come and hit each other and then they get stuck. We find in such a situation that mechanical energy is not conserved but momentum is conserved because there is no external force in the system. So, internal forces do cancel out and therefore, there is no there is definitely a conservation of momentum though not of energy. So, once we realize that in this particular case in the new definition in the relativity conservation of momentum critically depends on the conservation of the fourth component. Let us think whether this particular fourth component can be thought of somehow related to another important quantity which is called energy. This was my fourth component A4 was equal to m naught gamma u Ic. So, let us try to look whether this fourth component can be related to energy. That is what was the main role of Einstein to realize this particular component as the fourth component of course, historically that may not be the case of the concept of four vector probably came much later. Einstein had a different probably way of looking into this new definition of energy but now we can talk in this particular term. See energy must have dimension of energy obviously and we will realize that mass into velocity square is the definition is the dimension of energy. So, I write this A4 as m naught gamma u Ic then write E is equal to mc square because that will give me the dimension of energy. If I write this E is equal to mc square divide by c where m of course is defined as m naught gamma u then I can write the fourth component A4 as IE by c. Let me just take little bit of time to explain once more. So, I had A4 which I wrote as m naught gamma u Ic. Now, this m naught gamma u I call as m I write this is m Ic. Now, I want to call E is equal to mc square because that has the dimension of energy. So, I multiply this by c and divide this by c. So, this becomes I mc square by c and with this new definition this becomes IE by c. So, my fourth component now becomes related to the energy of the particle and is written as IE by c. Remember, we have used E is equal to mc square and m as m naught times gamma u. Remember, it is m naught which is a four scalar not this m because this m is m naught multiplied by gamma u and gamma u will be different in different frames because this depends on the speed of the particle and speed is indeed a frame dependent quantity. And as we have all known that this concept of new definition of energy E is equal to mc square is so popular that people know, people may not know physics, but they know E is equal to mc square. This has created huge impact on society just by the name of Einstein E is equal. We always associate with Einstein E is equal to mc square. It had a huge impact. So, what we said let us repeat now the conservation of momentum will be valid in all frames if energy E the way we have now defined is also conserved in the process. So, you must have both momentum and energy conservation unlike the case in classical mechanics where there are certain phenomenon in which we were not conserving the energy and in certain phenomena we were conserving energy. In fact strictly speaking in those phenomenon also the energy is conserved. Energy is not lost, but it is converted from the standard mechanical energy to some different form of energy, but strictly speaking the energy is not lost. It is always conserved, but now in relativity whether the type of collision is the one which we described earlier where the two particles come and get stuck to each other or whether it is what we traditionally called elastic collision. In all these cases we will conserve both energy and momentum if we want them to be universal phenomena. Now, we also see let us look at the fourth component of the transformation. The fourth component now I have replaced that gamma um with E prime. So, we have new definition of energy here. So, this is the transformation of the fourth component minus i beta gamma times the first component plus gamma times the fourth component. Now, if this is conserved this is conserved this automatically means this is also conserved. So, this ensures that even if energy that even energy E would be conserved in all frames if energy and momentum are conserved in a given frame. So, if momentum is conserved I need energy to be conserved if this momentum conservation has to be universal law, but it also assures that momentum and energy are conserved in one frame of a phrase both momentum and energy will be conserved in any other frame. Now, let us look at this particular energy what is this energy? See, we always think that we have always been telling that if you go to the non-relativistic limit it means if your particle speed is very small in comparison to speed of light then you must get back your classical mechanics that is what we always been telling. So, if my U was 0 for example, my gamma U will be equal to 1 the momentum because U is 0 here will be 0, but this energy will not be 0 because this becomes 1 this will be just m naught c square. Similarly, in the limit let us assume that U is not 0, but very small in comparison to speed of light. So, in the limit U being much smaller than c, gamma U will again 10 to 1 which means p naught will be approximately equal to m naught gamma m naught U because gamma U is more or less close to 1 which gives me back my classical definition of momentum which is mass multiplied by the speed or mass sorry mass multiplied by the velocity but my E will still remain m naught c square. So, we do not have a really classical analog of this particular energy, we had never thought that this if a particle is at rest when you have speed equal to 0 still the particle will have some energy. The classical mechanics we had not thought this is a totally new revolutionary concept of energy where we say that even if there is a particle which is at rest this particular particle has a rest mass energy that is what we call a rest mass energy of m naught c square. Now, the concept of energy is little more general in the eye of Einstein or in the eye of special theory of relativity it contains all the form of energy. If the particle gains speed its energy increases. So, this will be on the top of m naught c square. Similarly, if the particle becomes hot means it is gaining some energy its mass should increase its m naught should increase. So, it contains information about the entire energy of the system whatever might be the energy. For example, if you have two particles and these two particles have their own rest mass energies and they get bound then when they get bound certain amount of energy is released and that release of energy would eventually lead to a decrease in its rest mass because as far as relativity is concerned there is no difference between mass and energy because the two are related by a fundamental constant c square. So, a mass of the particle a rest mass of the particle particle will tell the entire energy contained in it. If you are talking for example, the rest mass of hydrogen atom the rest mass of hydrogen atom will be slightly smaller than the sum of the rest masses of proton and electron because in order to form a hydrogen atom certain amount of energy has been released and therefore the mass must have come down. Similarly, if I take hydrogen atom and release it so that we give it certain amount of energy so hydrogen so that the electron and the proton become separate then we will find that the masses of electron and proton where sum would be slightly larger than the mass of the hydrogen atom and that increase will be just related to the binding energy of the hydrogen atom. So, this is what I have written this new form of energy does not resemble any classically known form of energy it is entirely a new concept of energy. Energy gain in different form would lead to an increase in the mass. So, if we take one particular particle in principle in principle heat it then its mass should go up. Mass can be expressed in the unit of units of energy as the two are related through a fundamental constant. In fact, it is often told many of the physicists especially those who are working in the particle physics area they will always express mass in the units of energy. What is the mass of electron its 0.51 MeV. See MeV mega electron volt is the unit of energy but one says the mass is equal to energy. Similarly, what is the mass of the proton is approximately 940 MeV. Again we are expressing mass in terms of energy effectively means it is the value of m naught c square which is equal to that energy. So, there is no difference in relativity in the mass and energy because the two are related by of course, they are dimensionally different. No doubt they are different dimensionally but as far as the relationship is concerned they are related by a fundamental constant speed of light square. Now, let us talk of kinetic energy. This is something which we know in classical mechanics. So, let us define kinetic energy with this new definition of energy. As we have seen that a particle at rest also has energy which we call as rest mass energy. Now, if this particle particle has started moving then its energy would have gone up then whatever is the increase in the energy that is what we call as kinetic energy. See, if the particular particle has started moving then the energy will be given by gamma u m naught c square. When the particle was at rest this gamma u was equal to 1 and the energy was m naught c square. Once the particle has picked up certain speed the value of gamma u has gone up whatever might be the slight value but it has gone up and therefore, this energy e has increased. Whatever this increase in the energy e this was the final energy, this was the initial energy. This is what we will call kinetic energy as e 2 minus e 1 is equal to gamma u minus 1 m naught c square. We can also write because this is what we have defined as m. So, I can also call this as m c square minus m naught c square. So, kinetic energy in relativity is defined as either gamma u minus 1 m naught c square or as m c square minus m naught c square. We will just now show that this kinetic energy which will reduce to the traditional definition of kinetic energy in the limit of low speeds or low values of u. Let us see how. I have written gamma sorry k as m naught c square and this was the value of gamma u, gamma u is 1 upon under root 1 minus u square by c square. So, this is the value of m c square minus m naught c square. So, I take this particular quantity numerator. If I take this particular quantity numerator, I can write this as m naught c square multiplied by 1 minus u square by c square which is here to the power of minus half. This term remains identical which is minus m naught c square. Now, this particular thing because in the limit u is very small in comparison to c. This can be expanded into a series and you can retain just the first term and neglect higher order terms because u is very small in comparison to c. So, when I expand here, I will get this quantity multiplied by here because there is a minus half. So, this will become plus half. So, this will become approximately m naught c square multiplied by 1 plus. This sign becomes plus because of this negative sign here and there is a half. So, this becomes 1 plus u square by 2 c square under the limit that u is very small in comparison to c. This minus m naught c square is a kinetic energy. If you expand this, you are getting this m naught c square. This will cancel with this m naught c square. This c square in the second term would cancel with this c square and then you will just get half m naught u square which is the standard classical definition of kinetic energy. So, this new value, this new definition of kinetic energy does yield me to the classical expression in the classical limit. That is what we had expected. But let us realize that the e which we now call as total energy or to be more precise, total relativistic energy that does not have classical analog. That will not reduce to half m u square when the speed of particle is low. It is the kinetic energy which will reduce to half m naught u square. So, we should be differentiating the kinetic energy and the total relativistic energy. And remember, it is the total relativistic energy which must be conserved in a process along with the momentum. So, these are the new definitions just to summarize. p is equal to m u, e is equal to m c square, k is equal to m c square minus m naught c square. And m is defined as m naught divided by under root 1 minus u square by c square. Now, looking at these new definitions, let me rewrite my momentum energy four vector. We had earlier agreed that the first three components are p x, p y and p z. We have also discussed that the fourth component is i, e by c. So, what it means that if I find a particular particle in a frame of reference with the new definition of momentum and energy, its value of momentum or x component of momentum is p x, y component of momentum is p y, z component of momentum is p z and its energy is e, then these four components will transform to a different frame of reference as prime frame of reference. It means the new values in a different frame, the values of the x component of the momentum, y component of the momentum, z component of the momentum and the energy would be given by the same transformation equation, which we have used earlier. In fact, because it involves both momentum and energy, we generally call this as momentum energy four vector, not just the momentum four vector. So, if I have to transform the momentum, then all I have to do is to expand this particular matrix. So, I can find out that p x prime, for example, here p x prime will be equal to gamma times p x plus i beta gamma times i e by c, p y prime of course will be p y, p z prime will of course be equal to p z and i e prime by c will be equal to minus i beta gamma times p x plus gamma times i e by c. If I just make them just simplify this equation, this becomes my momentum energy transformation. So, p x prime would be given by gamma times p x minus v e by c square, p y prime becomes equal to p y, p z prime becomes equal to p z and e prime becomes gamma e minus v p x. Probably one would have noticed the symmetry, the arguments symmetry in these transformation equations. See, this type of equation x minus v t was appearing for x component. Here it now appears for the fourth component, which is the energy in the momentum, which was the first component. This equation is gamma p x minus v e by c square. Similar type of equation was for t, t prime was turning out to be equal to gamma t minus v x by c square. It is just to remember, but we realize their thing, which was the fourth component was totally different. Here the fourth component is i e by c. So, that makes a difference. There it was i c t. Here it is i e by c. So, that makes this particular difference. Now, let us look at the length of its energy momentum four vector for a single particle. See, in principle, we can write this particular momentum four vector for a set of n particles. We can add all these momentum. We will work out one or two examples for this particular case later, not in this lecture, but some of the later lectures. But at the moment, let us just assume that there is one single particle. And if it is so, then the length of this particular particle should be a four scalar. It means, even if I change my frame of reference, this should not change. It is true in principle for any four vector. Even if you would have written for the sum, the length should be frame independent quantity. But I am interested in looking at a specific expression, which eventually is obtained by obtaining the length of energy momentum four vector for a single particle. So, let us just evaluate the length of energy momentum four vector for just a single particle, which gives me one expression, which turns out to be a very useful expression for solving the problems. So, if I have to look at the length, what I have to do? I have to take the momentum four vector and take a dot product with its own. This will give me the square of the length of the energy momentum four vector. So, let us evaluate this p dot p, which I am doing in the next transparency. So, this was my p dot p. Remember, the first three components of p are px, py, pz and the fourth component was i, e, yc. Now, if I have to take a dot product, it means we take with its own self. So, I write the same vector, the component wise as i, e, yc. I take the dot product, it means this multiplied by this plus this multiplied by this plus this multiplied by this plus this multiplied by this. That is the way we have defined the dot product between two four vectors. So, if I multiply this by this, I get px square. So, this will turn out to be px square plus py square plus pz square because I have to add all these three. So, this multiplied by this plus this multiplied by this plus this multiplied by this plus this multiplied by this. This gives with px square plus py square plus pz square. When I multiply the two, i square will give me minus 1. So, this will be px square plus p y square plus pz square minus e square by c square. So, this is what I have written here, p dot p is equal to p square minus e square by c square. Now, we have already seen that the value of p is related to m naught gamma u u. So, ux square plus ui square plus uz square, I can write as u square. So, this becomes m naught square gamma u square u square and as far as e is concerned, this was also mc square and m was m naught gamma u. So, this becomes m naught square gamma u square c square. I can take this m naught square gamma u square out, common. So, this becomes m naught gamma u square. This becomes equal to u square minus c square. Now, you can very easily see that if I write this gamma u square and expand it, this will be 1 upon undone 1 minus u square by c square. If you just simplify this equation, which we have in one of the cases earlier, this will just become minus c square and this is minus m naught square c square, which indeed is a four scalar, because m naught we had defined as a four scalar. See, I know it is constant in all the frame of reference. So, the length of the four vector is indeed unchanged when I change my frame of reference. But this particular equation gives me something which is interesting. It tells me that for a single particle, p square minus e square by c square must be equal to minus m naught square c to the power 4. This is what I have written. This is obviously same for all the frames. This leads to the following useful relationship. e square is equal to p square c square plus m naught square c to the power 4. Let us just look it here. This was p square. So, what I take? I multiply by c square. Let me write it here. p square minus e square by c square is equal to minus m naught square c square. So, if I multiply the whole thing by c square, p square c square minus e square is equal to minus m naught, sorry there was m naught square here, m naught square c to the power 4. I take this on this side, this on this side. So, this becomes e square is equal to p square c square plus m naught square c to the power 4. This is a very useful equation, because this gives you relationship between energy and momentum. So, if I know the energy of the particle, I can find out what is the momentum of the particle. If I know the momentum of the particle, I can find out what is the energy of this particle. Hence, this particular equation is useful, because it gives me a relationship between energy and momentum. So, this is what is the equation, which we normally use in many of the equations requiring interaction of various particles giving me e square is equal to p square c square plus m naught square c to the power 4. Similarly, I can find out a relationship between kinetic energy and momentum. The kinetic energy has no very special meaning in relativity, but many times now we talk in terms of kinetic energy. So, if you are interested in finding out a relationship between kinetic energy and momentum, that also I can do. So, this becomes p square c square is equal to e square minus m naught square c to the power 4. The same equation I have written in a slightly different fashion. And this energy e, I have written as k plus m naught c square, because we had defined k as m c square minus m naught c square and e was m c square. So, this I can write as k plus m naught square, m naught c square whole square minus m naught square c to the power 4. You just open this up, this becomes k square plus 2 k m naught c square plus m naught square c to the power 4. That m naught square c to the power 4 cancels with this. This equation comes out to be equal to k square plus 2 m naught c square k. So, this is the relationship between kinetic energy and the momentum. Let us just look at the classical limit of this particular equation. The classical limit of this particular equation I just take k out. So, p square c square becomes k multiplied by k plus 2 m naught c square. So, if k is very small in comparison to the rest mass energy of the particle, I can neglect k. And if I neglect k, then I get k is equal to p is the c square cancels here. I get k is equal to p square by 2 m naught, which is the standard energy momentum relationship in classical mechanics or rather kinetic energy momentum relationship in the classical mechanics. p square by 2 m gives me the kinetic energy. This also tells you one more thing. Often people ask, when should we be sure that we should apply special theory of relativity without unless we will make a big mistake. And when we need not apply because we are really in a classical limit, in case of if we are talking about velocities, we always say that gamma should be very close to 1. It means you must be talking much less than 0.1 c of that order. In terms of energy, this always tells so long the kinetic energy of the particle is much smaller in comparison to the rest mass of energy of the particle than I can apply classical mechanics. For example, if I am talking of hydrogen atom where the particle has the energy of the order of 13 electron volt, we realize that it is rest mass energy is half MeV. Then of course, I can apply classical mechanics. The last thing which I would like to describe in today's lecture is a totally new concept which comes out of relativity, which is the presence of a zero rest mass particle. In classical mechanics, we never can imagine a particle which has no mass. But in relativistic mechanics, we can imagine that the particle particle may have zero rest mass. m naught can be 0. Of course, if m naught is equal to 0, using these two expressions for energy and momentum, energy should be 0 and p should be 0 provided this particular quantity is not 0. And this quantity will be 0 only when u becomes equal to c. So, this gives you a possibility that even if m naught is equal to 0, e can be finite non-zero, p can be finite non-zero. But in that case, you must have u equal to c. It means that particular particle must travel with speed of light. So, relativity gives you a possibility of presence of a zero rest mass particle, but that particle must travel with the speed of light. And we know a typical example is light itself, photon is considered as a particle which moves obviously with the speed of light and has zero rest mass. So, this is what I have to say unless u is equal to c. In such cases of course, because rest mass energy is 0. So, total energy is equal to the kinetic energy is equal to p c, because e square is equal to p square c square plus m naught square c to the power 4. That m naught square c to the power 4 term does not exist. So, it just becomes e is equal to p c. So, finally, I will summarize whatever we have discussed. We define energy momentum 4 vector in this particular lecture. Then, ensure that energy momentum conservation becomes indeed universal. And finally, give a totally new concept of zero rest mass particle. Thank you.